Given: x2+(p−3)x+p=0
∵ Roots are real and equal, then
b2−4ac=0
Here we compare the coefficients of a,b and c with the equation ax2+bx+c=0 .
a=1,b=p−3‌ and ‌c=p
Now putting the values of a,b and c in given equation
(p−3)2−4×1×p=0
p2+9−6p−4p=0
p2+9−10p=0
p2−10p+9=0
p2−9p−p+9=0
p(p−9)−1(p−9)=0
⇒‌‌(p−9)(p−1)=0
Hence, p=9 or 1